But since |x|=|x0+h|≦|x0|+|h| and also |x0|≦|x0|+|h|, so each summand in the parentheses is at most equal to (|x0|+|h|)n-1, and since there are n summands, the sum is at most equal to n⁢(|x0|+|h|)n-1. Thus we get

|Δ|≦n⁢|h|⁢(|x0|+|h|)n-1.

We may choose |h|<1; this implies

|Δ|≦n⁢|h|⁢(|x0|+1)n-1.

The right hand side of this inequality is less than ε as soon as we still require

|h|<εn⁢(|x0|+1)n-1.

This means that the power function x↦xn is continuous at the point x0.